{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

hmwk-6-K09-soln

# hmwk-6-K09-soln - PHY 317K Homework 6 Solutions 1 B For a...

This preview shows pages 1–3. Sign up to view the full content.

PHY 317K Homework 6 Solutions 1. B For a circle (like a coin), A = πr 2 = πd 2 / 4, where d is the diameter. If d changes by Δ d , the area becomes A + Δ A = π ( d + Δ d ) 2 / 4 = πd 2 / 4 + 2 πd Δ d/ 4 + π d ) 2 / 4. Therefore, Δ A = 2 πd Δ d/ 4, if we ignore the third term since Δ d is very small compared to d . Then if we multiply the left side by d/d , we see Δ A = 2( πd 2 / 4)Δ d/d , from which it follows: Δ A/A = 2Δ d/d . Hence, a 0.17% change in d results in a 0.34% change in the area of the coin. 2. C For coefficient of linear expansion α , Δ V V = 3 α Δ T . Thus, here Δ V = 3 αV Δ T = 3(1 . 1 × 10 - 5 /C )(4 . 00 cm) 3 (57 C ) = 0 . 12 cm 3 . 3. D The heat capacity at constant volume and the heat capacity at constant pressure have different values because the system does work at constant pressure, but not at constant volume. I.e., when the volume changes, work is done. 4. A The energy needed to melt A is ML , where M is the mass of A and L is latent heat of fusion. The total heat energy that A receives is C A ( T final - T A ) + ML , and this must equal the heat energy lost by B : C B ( T B - T final ) = C A ( T final - T A ) + ML. Solving for T final we find: T final = C A T A + C B T B - ML C A + C B 5. A During the time that latent heat is involved in a change of phase the temperature does not change. 6. E Several steps are involved: 1) The ice is raised from - 20 C to 0 C: E 1 = 10 g(0 . 5 cal / g · C )(20 C ) = 100 cal. 2) The ice is melted: E 2 = 10 g(80 cal / g) = 800 cal. 3) The water is heated to 100 C: E 3 = 10 g(1 . 00 cal / g · C )(100 C ) = 1000 cal. 4) The water is vaporized: E 4 = 10 g(540 cal / g) = 5400 cal. 5) The steam is heated to 130 C: E 5 = 10 g(0 . 5 cal / g · C )(60 C ) = 300 cal. The sum of the energies required for all steps is 7600 cal. 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
7. C The humidifier heats 15 g of water from 20 C to 100 C and then evaporates those 15 g each minute. The energy needed for that is: E = mc Δ T + mL V = (15 g) · (4190 J / kg K) · (80 K) + (15 g) · (2256 kJ / kg) = 39000 J where L V is heat of vaporization. 8. B According to the first law of thermodynamics, applied to gas, the increase in the internal energy during the process is equal to the heat input plus the work done on the gas. 9. D The work done by the gas is equal to W = pdV . Because of the direction of the processes, we see that I and II result in negative work, while III, IV, and V result in positive work. In magnitude, the positive work in IV is largest. This is simply based on the area under the curves.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}